If $(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = x + i y$ . Then $2.5.10....... (1 + n^{2}) = x^{2} + y^{2}$

True

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False

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Solution

The correct option is A True
GIven that $(1 + i) (1 + 2 i) (1 + 3 i) . . . (1 + n i) = x + i y$

on computing the mod values we get

$\sqrt{(1^{2} + 1^{2}) (1^{2} + 2^{2}) (1^{2} + 3^{2}) . . . (1^{2} + n^{2})} = \sqrt{x^{2} + y^{2}}$

$\Rightarrow (2) (5) (10) . . . (1 + n^{2}) = x^{2} + y^{2}$

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Similar questions

(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = x + i y

2.5.10..... (1 + n^{2}) = x^{2} + y^{2}

(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = α + i β

2.5.10 . . . (1 + n^{2}) =

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+ $n^{2}$ ) is equal to

(1 - i) (1 - 2 i) (1 - 3 i) . . . (1 - n i) = x - i y

2 \cdot 5 \cdot 10..... (1 + n^{2}) =

$If x + i y = (1 + i) (1 + 2 i) (1 + 3 i), then x^{2} + y^{2}$

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